Usually, blurring filters have the sum of all the values in the filter matrix equal to $1$.
Why is it so?
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$\begingroup$ The question contradicts the example, if the matrix that is provided is supposed to be a 3x3 convolution matrix. Can you please clarify? $\endgroup$ – A_A Oct 5 '18 at 12:25
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Blurring an image means reducing its high frequencies while retaining its low frequencies.
Ususally this means a lowpass filter with a cutoff frequency of $\omega_c$. A standard low pass filter would have a DC response of $1$; i.e., $$H(0,0) = 1$$
This translates into time domain using the DTFT as: $$ H(0,0) = 1 \implies \sum_n \sum_m h[n,m]e^{-j0n}e^{-j0n} =\sum_n \sum_m h[n,m] = 1 $$
Which indicates that the sum of the impulse response samples equates to $1$.
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For constant images to remain constant after blurring.
The most blurred images are flat-valued with any constant $c$. They are left invariant by traditional smoothers, low-pass or blur filters, hence $\sum h_{k,l}c=c$, for all $c\neq 0$ when $\sum h_{k,l}=1$,
answered Oct 5 '18 at 15:33
